Lecture Notes on Analysis II MA131. Xue-Mei Li

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1 Lecture Notes on Analysis II MA131 Xue-Mei Li March 8, 2013

2 The lecture notes are based on this and previous years lectures by myself and by the previous lecturers. I would like to thank David Mond, who is the driving force for me to finish this work and for his important feed backs. In this edition I would like to thank Ruoran Huang and Chris Midgley for a list of typos. Xue-Mei Li Office C Important note: Please tell Xue-Mei Li or David Mond if you find any mistakes in these lecture notes, so that future printed editions, and the online lecture notes, can be improved. If something seems like a mistake but you are not sure, write to us anyway if there is a mistake, your message will help correct it, and if there is not, you will learn something about Analysis. 1

5 10 Classical Functions The Exponential and the Natural Logarithm Function The Sine and Cosine Functions Taylor s Theorem and Taylor Series Non-centered Power Series Uniqueness of Power Series Expansion Taylor s Theorem with remainder in Lagrange form Taylor s Theorem in Integral Form* Cauchy s Mean Value Theorem A Table Techniques for Evaluating Limits Use of Taylor s Theorem L Hôpital s Rule Unfinished Business 160 4

6 The module 0.1 Introduction Analysis II, together with Analysis I, is a 24 CATS core module for first year students. The content of the module will be delivered in 29 lectures. Assessment 7.5%, Term 1 assignments, 7.5%, Term 2 assignments, 25%, January exam (on Analysis 1) 60%, final exam (3 hour in June, cover analysis 1+2) Fail the final ===> Fail the module. Assignments: Given out on Thursday and Fridays in lectures, to hand in the following Thursday in supervisor s pigeonhole by 14:00. Each assignment is divided into part A, part B and part C. Part B is for assignments. Learn Definitions and statements of theorems. Take the exercises very seriously. It is the real way you learn (and hence pass the summer exam) Books: G. H. Hardy: A Course of Pure Mathematics, Third Edition (First Edition 1908), Cambridge University Press. E. T. Whittaker &G. N. Watson A Course of Modern Analysis T. M. Apostol Calculus, Vol I&II, 2nd Edition, John Wiley & Sons E. Hairer & G. Wanner: Analysis by Its History, Springer. Objectives of the module: Study functions of one real variable. This module leads to analysis of functions of two or more variables (Analysis III), Ordinary differential equations, partial differential equations, Fourier Analysis, Functional analysis etc. What we cover: 5

7 Continuity of functions of one real variable (lectures 1-3) the Intermediate Value theorem, continuous functions on closed intervals Continuous limits (circa lecture 8-9) Extreme Value Theorem, Inverse Function Theorem (circa 10-11) Differentiation (circa lecture 12), Mean Value Theorem Power series (ca. lecture 23) Taylor s Theorem (ca. lecture 23) L Hôpital s rule (ca. lecture 26) Relation to Analysis I. Continuity and continuous limits of function will be formulated in terms of sequential limits. Power series are functions obtained by sums of infinite series. Relation to Analysis II. Uniform convergence and the Fundamental Theorem of Calculus from Analysis III would make a lot of proofs here easier. Relation to Complex Analysis The power series expansion. Relation to Ordinary Differential Equations Differentiability of functions. Solve e.g. f (x) = f(x), f(0) = 1. Feedbacks Ask questions in lectures. Talk to me after or before the lectures. Plan your study: Shortly after each hour of lecture, you need to spend one hour going over lecture notes and mulling over the assignments. It is a rare student who has understood everything during the lecture. There is plentiful evidence that putting in this effort shortly after the lecture pays ample dividends in terms of understanding. In particular, if you begin a lecture having understood the previous one, you learn much more from it, so the process is cumulative. Please plan two hours per week for the exercises. 6

10 Chapter 1 Continuity of Functions of One Real Variable Let R be the set of real numbers. We will often use the letter E to denote a subset of R. Here are some examples of the kind of subsets we will be considering: E = R, E = (a, b) (open interval), E = [a, b] (closed interval), E = (a, b] (semi-closed interval), E = [a, b), E = (a, ), E = (, b), and E = (1, 2) (2, 3). The set of rational numbers Q is also a subset of R. 1.1 Functions Lecture 1 Definition By a function f : E R we mean a rule which to every number in E assigns a number from R. This correspondence is denoted by y = f(x), or x f(x). We say that y is the image of x and x is a pre-image of y. The set E is the domain of f. The range, or image, of f consists of the images of all points of E. It is often denoted f(e). We denote by N the set of natural numbers, Z the set of integers and Q the set of rational numbers: N = {1, 2,... } Z = {0, ±1, ±2,... } Q = { p q : p, q Z, q 0}. 9

12 and by symmetry a b + b a. Pythagorean theorem : sin 2 x + cos 2 x = 1. cos 2 x sin 2 x = cos(2x), cos(x) = 1 2 sin 2 ( x 2 ). 1.3 Continuous Functions Lecture 2 What do we mean by saying that f(x) varies continuously with x? It is reasonable to say f is continuous if the graph of f is an unbroken continuous curve. The concept of an unbroken continuous curve seems easy to understand. However we may need to pay attention. For example we look at the graph of { x, x 1 f(x) = x + 1, if x > 1 It is easy to see that the curve is continuous everywhere except at x = 1. The function is not continuous at x = 1 since there is a gap of 1 between the values of f(1) and f(x) for x close to 1. It is continuous everywhere else. y 1 x Now take the function F (x) = { x, x 1 x , if x > 1 The function is not continuous at x = 1 since there is a gap of However can we see this gap on a graph with our naked eyes? No, unless you have exceptional eyesight! 11

13 Here is a theorem we will prove, once we have the definition of continuous function. Theorem (Intermediate value theorem): Let f : [a, b] R be a continuous function. Suppose that f(a) f(b), and that the real number v lies between f(a) and f(b). Then there is a point c [a, b] such that f(c) = v. This looks obvious, no? In the picture shown here, it says that if the graph of the continuous function y = f(x) starts, at (a, f(a)), below the straight line y = v and ends, at (b, f(b)), above it, then at some point between these two points it must cross this line. y f(b) v f(a) y=f(x) a c b x But how can we prove this? Notice that its truth uses some subtle facts about the real numbers. If, instead of the domain of f being an interval in R, it is an interval in Q, then the statement is no longer true. For example, we would probably agree that the function F (x) = x 2 is continuous (soon we will see that it is). If we now consider the function f : Q Q defined by the same formula, then the rational number v = 2 lies between 0 = f(0) and 9 = f(3), but even so there is no c between 0 and 3 (or anywhere else, for that matter) such that f(c) = 2. Sometimes what seems obvious becomes a little less obvious if you widen your perspective. 12

14 These examples call for a proper definition for the notion of continuous function. In Analysis, the letter ε is often used to denote a distance, and generally we want to find way to make some quantity smaller than ε : The sequence (x n ) n N converges to l if for every ε > 0, there exists N N such that if n > N, then x n l < ε. The sequence (x n ) n N is Cauchy if for every ε > 0, there exists N N such that if m, n > N then x m x n < ε. In each case, by taking n, or n and m, sufficiently big, we make the distance between x n and l, or the distance between x n and x m, smaller than ε. Important: The distance between a and b is a b. The set of x such that x a < δ is the same as the set of x with a δ < x < a δ. The definition of continuity is similar in spirit to the definition of convergence. It too begins with an ε, but instead of meeting the challenge by finding a suitable N N, we have to find a positive real number δ, as follows: Definition Let E be subset of R and c a point of E. 1. A function f : E R is continuous at c if for every ε > 0 there exists a δ > 0 such that if x c < δ and x E then f(x) f(c) < ε. 2. If f is continuous at every point c of E, we say f is continuous on E or simply that f is continuous. The reason that we require x E is that f(x) is only defined if x E! If f is a function with domain R, we generally drop E in the formulation above. Example The function f : R R given by f(x) = 3x is continuous at every point x 0. It is very easy to see this. Let ε > 0. If δ = ε/3 then x x 0 < δ = x x 0 < ε/3 = 3x 3x 0 < ε = f(x) f(x 0 ) < ε as required. Exercise For each of the following cases, show that the function given is continuous at every point x 0 R. 13

15 1. f(x) = 3x f(x) = 5x 3. f(x) = f(x) = λ with λ a non-zero constant. Example Prove that the function f : R R given by f(x) = x 2 + x is continuous at x = 2. Take ε > 0. We wish to have f(x) f(2) < ɛ. Let us simplify and bound the left hand side quantity, f(x) f(2) = x 2 + x (4 + 2) x x 2 = x + 2 x 2 + x 2 = ( x ) x 2. Now we should give an upper bound for x + 2. If x 2 1, then This shows that x + 2 = (x 2) + 4 x f(x) f(2) ( x ) x 2 6 x 2. We can take δ = min( ε 6, 1). If x 2 < δ, the above reasoning gives f(x) f(2) < ɛ. There are two things which make the proof easier than it looks at first sight: 1. If δ works, then so does any δ > 0 which is smaller than δ. There is no single right answer. Sometimes you can make a guess which may not be the biggest possible δ which works, but still works. In the last example, we could have taken δ = min{ ε 10, 1} instead of min{ ε 6, 1}, just to be on the safe side. It would still have been right. 14

16 2. Fortunately, we don t need to find δ very often. In fact, it turns out that sometimes it is easier to use general theorems than to prove continuity directly from the definition. Sometimes it s easier to prove general theorems than to prove continuity directly from the definition in an example. My preferred proof of the continuity of the function f(x) = x 2 + x goes like this: (a) First, prove a general theorem showing that if f and g are functions which are continuous at c, then so are the functions f + g and fg, defined by { (f + g)(x) = f(x) + g(x) (fg)(x) = f(x) g(x) (b) Second, show how to assemble the function h(x) = x 2 + x out of simpler functions, whose continuity we can prove effortlessly. In this case, h is the sum of the functions x x 2 and x x. The first of these is the product of the functions x x and x x. Continuity of the function f(x) = x is very easy to show! As you can imagine, the general theorem in the first step will be used on many occasions, as will the continuity of the simple functions in the second step. So general theorems save work. But that will come later. For now, you have to learn how to apply the definition directly in some simple examples. First, a visual example with no calculation at all: 15

17 f(c) + ε f(c) f(c) ε a c b Here, as you can see, for all x in (a, b), we have f(c) ε < f(x) < f(c)+ε. What could be the value of δ? The largest possible interval centred on c is shown in the picture; so the biggest possible value of δ here is b c. If we took as δ the larger value c a, there would be points within a distance δ of c, but for which f(x) is not within a distance ε of f(c). Exercise Suppose that f(x) = x 2, x 0 = 7 and ε = 5. What is the largest value of δ such that if x x 0 < δ then f(x) f(x 0 ) < ε? What if ε = 50? What if ε = 0.1? Drawing a picture like the last one may help. Discontinuity Let us now do a logic problem. We gave a definition of what is meant by f is continuous at c : ε > 0 δ > 0 such that if x c < δ then f(x) f(c) < ε. How to express the meaning of f is not continuous at c in these same terms? there exists an ε > 0 such that for all δ > 0, there exists an x satisfying x c < δ but f(x) f(c) ε. 16

18 Example Consider f : R R, { 0, if x Q f(x) = 1, if x Q. Then f is discontinuous at every point. The key point is that between any two numbers there is an irrational number and a rational number. More precisely, take ε = 0.5. Let δ > 0 be any positive number. If c Q, then f(c) = 0. There is an x Q with x c < δ; because x Q, f(x) = 1. Thus f(x) f(c) = 1 0 > 0.5. So f is not continuous at c. If c Q then f(c) = 1. Take ε = 0.5. For any δ > 0, take x Q with x c < δ; then f(x) f(c) = 0 1 > 0.5 = ε. Example Consider f : R R, { 0, if x Q f(x) = x, if x Q Claim: This function is continuous at c = 0 and discontinuous everywhere else. Let c = 0. For any ε > 0 take δ = ε. If x c < δ, then f(x) f(0) = x in case c Q, and f(x) f(0) = 0 0 = 0 in case c Q. In both cases, f(x) f(0) x < δ = ε. Hence f is continuous at 0. Exercise Show that the function f of the last example is not continuous at c 0. Hint: take ε = c /2. Example Suppose that g : R R and h : R R are continuous, and let c be some fixed real number. Define a new function f : R R by { g(x), x < c f(x) = h(x), x c. Then the function f is continuous at c if and only if g(c) = h(c). Proof: Since g and h are continuous at c, for any ε > 0 there are δ 1 > 0 and δ 2 > 0 such that g(x) g(c) < ε when x c < δ 1 and such that h(x) h(c) < ε when x c < δ 2. 17

20 Example Let f : R R. { 1/q, if x = p f(x) =, q > 0 q 0, if x Q p, q coprime integers Then f is continuous at all irrational points, and discontinuous at all rational points. Proof Every rational number p/q can be written with positive denominator q, and in this proof we will always use this choice. Case 1. Let c = p/q Q. We show that f is not continuous at c. Take ε = 1 2q. No matter how small is δ there is an irrational number x such that x c < δ. And f(x) f(c) = 1 q > ε. Case 2. Let c Q. We show that f is continuous at c. Let ε > 0. If x = p Q, q f(x) f(c) = 1. So the only rational numbers p/q for which f(x) f(p/q) ε q are those with denominator q less than or equal to 1/ε. Let A = {x Q (c 1, c + 1) : q 1 }. Clearly c / A. The crucial point is ε that if it is not empty, A contains only finitely many elements. To see this, observe that its members have only finitely many possible denominators q, since q must be a natural number 1/ε. For each possible q, we have p/q (c 1, c + 1) p (qc q, qc + q), so no more than 2q different values of p are possible. It now follows that the set B := { x c : x A}, if not empty, has finitely many members, all strictly positive. Therefore if B is not empty, it has a least element, and this element is strictly positive. Take δ to be this element. If B is empty, take δ =. In either case, (c δ, c + δ) does not contain any number from A. Suppose x c < δ. If x / Q, then f(x) = f(x) = 0, so f(x) f(c) = 0 < ε. If x = p/q Q then since x / A, f(x) f(c) = 1 q < ε. Example f : ( 1, 1) R. f(x) = ( 1) n xn n n=1 Is f continuous at 0? This is a difficult problem for us at the moment. We cannot even graph this function and so we have little intuition. But we can answer this question after the study of power series. Exercise Is there a value a such that the function 1, if x > 0 f a (x) = a, if x = 0 1, if x < 0. is continuous at x = 0? Justify your answer. 19

21 1.4 Continuity and Sequential Continuity Lecture 3 Definition We say f : E R is sequentially continuous at c E if for every sequence {x n } E such that lim n x n = c we have lim f(x n) = f(c). n Theorem Let E be a subset of R and c E. Let f : E R. The following are equivalent: 1. f is continuous at c. 2. f is sequentially continuous at c. Proof Suppose that f is continuous at c. For any ε > 0 there is a δ > 0 so that f(x) f(c) < ε, whenever x c < δ. Let {x n } be a sequence with lim n x n = c. There is an integer N such that x n c < δ, n > N, Hence if n > N, f(x n ) f(c) < ε. We have proved that f(x n ) f(c). Thus f is sequentially continuous at c. Suppose that f is not continuous at c. Then ε > 0 such that for every δ > 0, there is a point x E with x c < δ but f(x) f(c) ε. In particular, this holds for δ = 1 n : there exists a point x n E with x n c < 1 n, but f(x n ) f(c) ε. In this way we have constructed a sequence {x n } which tends to c, but for which f(x n ) f(c). This shows that f is not sequentially continuous at c. 20

22 Sequential continuity is handy when it comes to showing a function is discontinuous. Example Let f : R R be defined by { sin( 1 f(x) = x ) x 0 0 x = 0 is not continuous at 0. Proof Observe that f(0) = 0. Take x n = 1 2nπ+ π ; then x n 0 as n, 2 but f(x n ) = 1 0. Hence f is not sequentially continuous at 0 and it is therefore not continuous at 0. Example Let f(x) = { x + 1 x 2 12 x = 2 The function f is not continuous at x = 2. Proof Take x n = n ; then x n 2 as n, but f(x n ) = n 3 f(2). Hence f is not sequentially continuous at 2 and it is therefore not continuous at 2. We end this section with a small result which will be very useful later. The following lemma says that if f is continuous at c and f(c) 0 then f does not vanish close to c. In fact f has the same sign as f(c) in a neighbourhood of c. We introduce a new notation: if c is a point in R and r is also a real number, then B r (c) = {x R : x c < r}. It is sometimes called the ball of radius r centred on c. The same definition makes sense in higher dimensions (B r (c) is the set of points whose distance from c is less than r); in R 2 and R 3, B r (c) looks more round than it does in R. Lemma Non-vanishing lemma Suppose that f : E R is continuous at c. 21

24 What can we deduce about the continuity of f + g, fg and f/g if f and g are continuous? Proposition The algebra of continuous functions Suppose f and g are both defined on a subset E of R and are both continuous at c, then 1. af + bg is continuous at c where a and b are any constants. 2. fg is continuous at c 3. f/g is continuous at c if g(c) 0. Note: If g(c) 0 then there is a neighbourhood of c on which g(x) 0, by the non-vanishing lemma The function f/g referred to in part (3) of the proposition is well defined on this neighbourhood, and it is this function, on this neighbourhood, that we claim is continuous. Proof Let {x n } be any sequence in E converging to c. Then lim f(x n) = f(c), n lim g(x n) = g(c). n By the algebra of convergent sequences we see that lim [af(x n) + bg(x n )] = [af(c) + bg(c)]. n Thus lim (af + bg)(x n) = (af + bg)(c). n Hence af + bg is sequentially continuous at c. Since lim n f(x n )g(x n ) = f(c)g(c), lim (fg)(x n) = (fg)(c) n and fg is sequentially continuous at c. Suppose that g(c) 0. By Lemma 1.4.5, there is a neighbourhood B r (c) such that g(x) 0. We may now assume that E = B r (c) and hence assume that x n B r (c). Any sequences in B r (c) satisfies g(x n ) 0. By sequential continuity of g, lim n g(x n ) = g(c) 0. Consequently f(x n ) lim n g(x n ) = f(c) g(c) and f g is sequentially continuous at c. 23

25 The continuity of the function f(x) = x 2 + x, which we proved from first principles in Example 1.3.5, can be proved more easily using Proposition For the function g(x) = x is obviously continuous (take δ = ε), the function h(x) = x 2 is equal to g g and is therefore continuous by 1.5.2(2), and finally f = h + g and is therefore continuous by 1.5.2(1). Example The function f given by f(x) = 1 x is continuous on R {0}. It cannot be extended to a continuous function on R since if x n = 1 n, f(x n ). Proposition composition of continuous functions Suppose f : E R is continuous at c, g is defined on the range of f, and g is continuous at f(c). Then g f is continuous at c. Proof Let {x n } be any sequence in E converging to c. By the continuity of f at c, lim n f(x n ) = f(c). By the continuity of g at f(c), lim g(f(x n)) = g(f(c)). n This shows that g f is continuous at c. Example A polynomial of degree n is a function of the form P n (x) = a n x n + a n 1 x n a 0. Polynomials are continuous, by repeated application of Proposition 1.5.2(1) and (2). 2. A rational function is a function of the form: P (x) Q(x) where P and Q are polynomials. The rational function P Q is continuous at x 0 provided that Q(x 0 ) 0, by 1.5.2(3). Example The exponential function exp(x) = e x is continuous. This we will prove later in the course. 2. Given that exp is continuous, the function g defined by g(x) = exp(x 2n+1 + x) is continuous (use continuity of exp, 1.5.5(1) and 1.5.4). Example The function x sin x is continuous. This will be proved later using the theory of power series. An alternative proof is given in the next section. From the continuity of sin x we can deduce that cos x, tan x and cot x are continuous: 24

26 Example The function x cos x is continuous by 1.5.4, since cos x = sin(x + π 2 ) and is thus the composite of the continuous function cos and the continuous function x x + π/2 Example The function x tan x is continuous. Use tan x = sin x cos x and 1.5.2(3). Discussion on tan x. If we restrict the domain of tan x to the region ( π 2, π 2 ), its graph is a continuous curve and we believe that tan x is continuous. On a larger range, tan x is made of disjoint pieces of continuous curves. How could it be a continuous function? Surely the graph looks discontinuous at π 2!!! The trick is that the domain of the function does not contain these points where the graph looks broken. By the definition of continuity we only consider x with values in the domain. The largest domain for tan x is R { π 2 + kπ, k Z} = k Z(kπ π 2, kπ + π 2 ). For each c in the domain, we locate the piece of continuous curve where it belongs. We can find a small neighbourhood on which the graph is is part of this single piece of continuous curve. For example if c ( π 2, π 2 ), make sure that δ is smaller than min( π 2 c, c + π 2 ). 1.6 Continuity of Trigonometric Functions Measurement of angles Two different units are commonly used for measuring angles: Babylonian degrees and radians. We will use radians. The radian measure of an angle x is the length of the arc of a unit circle subtended by the angle x. x 1 x 25

27 The following explains how to measure the length of an arc. Take a polygon of n segments of equal length inscribed in the unit circle. Let l n be its length. The length increases with n. As n, it has a limit. The limit is called the circumference of the unit circle. The circumference of the unit circle was measured by Archimedes, Euler, Liu Hui etc.. It can now be shown to be an irrational number. Historically sin x and cos x are defined in the following way. Later we define them by power series. The two definitions agree. Take a right-angled triangle, with the hypotenuse of length 1 and an angle x. Define sin x to be the length of the side facing the angle and cos x the length of the side adjacent to it. Extend this definition to [0, 2π]. For example, if x [ π 2, π], define sin x = cos(x π 2 ). Extend to the rest of R by decreeing that sin(x + 2π) = sin x, cos(x + 2π) = cos x. Lemma If 0 < x < π 2, then sin x < x < tan x. Proof Take a unit disk centred at 0, and consider a sector of the disk of angle x which we denote by OBA. y B O F A E x x 2π The area of the sector OBA is: times the area of the disk which is π. hence Area(Sector OBA) = 1 2 x. Let a = OF and b = F A then a + b = 1. The area of the triangle OBA is the sum of the areas of OBF and BF A: Area(triangle OBA) = 1 2 a sin(x) b sin x = 1 2 sin(x). 26

30 Chapter 2 Continuous Functions on closed Intervals: I Definition A number c is an upper bound of a set A if for all x S we have x c. A number c is the least upper bound of a set A if c is an upper bound for A. if U is an upper bound for A then c U. The least upper bound of the set A is denoted by sup A. The completeness Axiom If S R is a non empty set bounded above it has a least upper bound in R. In particular there is a sequence x n S such that c 1 n x n < c. This sequence x n converges to c. 2.1 The Intermediate Value Theorem Lectures 5 & 6 Recall Theorem 1.3.1, which we state again: Theorem Intermediate Value Theorem (IVT) Let f : [a, b] R be continuous. Suppose that f(a) f(b). Then for any v strictly between f(a) and f(b), there exists c (a, b) such that f(c) = v. A rigorous proof was first given by Bolzano in Proof We may assume f(a) < f(b) (in case f(b) < f(a) define g = f.) Consider the set A = {x [a, b] : f(x) v}. Note that a A and A is 29

31 bounded above by b, so it has a least upper bound, which we denote by c. We will show that f(c) = v. y f v a c b x Since c = sup A, there exists x n A with x n c. Then f(x n ) f(c) by continuity of f. Since f(x n ) v then f(c) v. Suppose f(c) < v. Then by the Non-vanishing Lemma applied to the function x v f(x), there exists r > 0 such that for all x B r (c), f(x) < v. But then c + r/2 A. This contradicts the fact that c is an upper bound for A. The idea of the proof is to identify the greatest number in (a, b) such that f(c) = v. The existence of the least upper bound (i.e. the completeness axiom) is crucial here, as it is on practically every occasion on which one wants to prove that there exists a point with a certain property, without knowing at the start where this point is. Exercise To test your understanding of this proof, write out the (very similar) proof for the case f(a) > v > f(b). Example Let f : [0, 1] R be given by f(x) = x 7 +6x+1, x [0, 1]. Does there exist x [0, 1] such that f(x) = 2? Proof The answer is yes. Since f(0) = 1, f(1) = 8 and 2 (1, 8), there is a number c [0, 1] such that f(c) = 2 by the IVT. 30

32 Remark*: There is an alternative proof for the IVT, using the bisection method. It is not covered in lectures. Suppose that f(a) < 0, f(b) > 0. We construct nested intervals [a n, b n] with length decreasing to zero. We then show that a n and b n have common limit c which satisfy f(c) = 0. Divide the interval [a, b] into two equal halves: [a, c 1], [c 1, b]. If f(c 1) = 0, done. Otherwise on (at least) one of the two sub-intervals, the value of f must change from negative to positive. Call this subinterval [a 1, b 1]. More precisely, if f(c 1) > 0 write a = a 1, c 1 = b 1; if f(c 1) < 0, let a 1 = c 1, b 1 = b. Iterate this process. Either at the k th stage f(c k ) = 0, or we obtain a sequence of intervals [a k, b k ] with [a k+1, b k+1 ] [a k, b k ], a k increasing, b k decreasing, and f(a k ) < 0, f(b k ) > 0 for all k. Because the sequences a k and b k are monotone and bounded, both converge, say to c and c respectively. We must have c c, and f(c ) 0 f(c ). If we can show that c = c then since f(c ) 0, f(c ) 0 then we must have f(c ) = 0, and we have won. It therefore rermains only to show that c = c. I leave this as an (easy!) exercise. Example The polynomial p(x) = 3x 5 + 5x + 7 = 0 has a real root. Proof Let f(x) = 3x 5 + 5x + 7. Consider f as a function on [ 1, 0]. Then f is continuous and f( 1) = = 1 < 0 and f(0) = 7 > 0. By the IVT there exists c ( 1, 0) such that f(c) = 0. Exercise Show that every polynomial of odd degree has a real root. Discussion on assumptions. In the following examples the statement fails. For each one, which condition required in the IVT is not satisfied? Example Let f : [ 1, 1] R, { x + 1, x > 0 f(x) = x, x < 0 Then f( 1) = 1 < f(1) = 2. Can we solve f(x) = 1/2 ( 1, 2)? No: that the function is not continuous on [ 1, 1]. 2. Define f : Q [0, 2] R by f(x) = x 2. Then f(0) = 0 and f(2) = 4. Is it true that for each v with 0 < v < 4, there exists c Q [0, 2] such that f(c) = v? No! If, for example, v2, then there is no rational number c such that f(c) = v. Here, the domain of f is Q [0, 2], not an interval in the reals, as required by the IVT. 31

33 Example The Intermediate Value Theorem may hold for some functions which are not continuous. For example it holds for the following function: { sin(1/x), x 0 f(x) = 0 x = 0, 0 x 1. Imagine the graphs of two continuous functions f and g. If they cross each other where x = c then f(c) = g(c). If the graph f is higher at a and the graph of g is higher at b then they must cross, as is shown below. Example Let f, g : [a, b] R be continuous functions. Suppose that f(a) < g(a) and f(b) > g(b). Then there exists c (a, b) such that f(c) = g(c). Proof Define h = f g. Then h(a) < 0 and h(b) > 0 and h is continuous. Apply the IVT to h: there exists c (a, b) with h(c) = 0, which means f(c) = g(c). Let f : E R be a function. Any point x E such that f(x) = x is called a fixed point of f. Theorem (Fixed Point Theorem) Suppose g : [a, b] [a, b] is a continuous function. Then there exists c [a, b] such that g(c) = c. Proof The notation that g : [a, b] [a, b] implies that the range of f is contained in [a, b]. Set f(x) = g(x) x. Then f(a) = g(a) a a a = 0, f(b) = g(b) b b b = 0. If f(a) = 0 then a is the sought after point. If f(b) = 0 then b is the sought after point. If f(a) > 0 and f(b) < 0, apply the intermediate value theorem to f to see that there is a point c (a, b) such that f(c) = 0. This means g(c) = c. Remark This theorem has a remarkable generalisation to higher dimensions, known as Brouwer s Fixed Point Theorem. Its statement requires the notion of continuity for functions whose domain and range are of higher dimension - in this case a function from a product of intervals [a, b] n to itself. Note that [a, b] 2 is a square, and [a, b] 3 is a cube. I invite you to adapt the definition of continuity we have given, to this higher-dimensional case. 32

34 Theorem (Brouwer, 1912) Let f : [a, b] n [a, b] n be a continuous function. Then f has a fixed point. The proof of Brouwer s theorem when n > 1 is rather harder than when n = 1. It uses the techniques of Algebraic Topology. Exercises Suppose that a sequence (x n ) is defined by a starting with some initial value x 1, choosing a function f, and for n 2 taking x n = f(x n 1 ) (Question 2 contains some examples). Show that if lim n x n = l, and if f is continuous at l, then f(l) = l. 2. Find the limit of the sequences defined by (a) x 1 = 1, x n+1 = 1 1+x n for n 0. (b) x 1 = 2, x n+1 = 1 2+x n for n A sequence of rectangles (R n ) is defined as follows: beginning with some rectangle R 1, at each stage we construct R n+1 by by adding to R n a square drawn on its longer side. H G A D F B C E R1=ABCD R2=ABEF R3=HBEG To do: (a) Suppose that the sequence has a limiting shape (i.e. sequence length of long side x n := length of short side that the tends to a limit). Show that this limit has only one possible value, and find it. (b) Show that the sequence does have a limiting shape. 33

35 Chapter 3 Continuous Limits 3.1 Continuous Limits Lecture 7 We wish to give a precise meaning to the statement f approaches l as x approaches c which is denoted by If f(x) = lim f(x) = l. x c { 2x + 1 x 1 0, x = 1 Could we say something about f(x) as x approaches 1? Let us agree that there is a limit. Should the limit be 3 or 0? It should be 3, and the value f(1) = 0 is not relevant. We do not care the value of f at c, and indeed for that matter f is not required to be defined at c. Definition Let c (a, b). Let l be a real number. Let f : E R. (1) Suppose that E (a, c) (c, b). We say that f tends to l as x approaches c, and write lim f(x) = l, x c if for any ε > 0 there exists δ > 0, such that for all x E satisfying we have 0 < x c < δ (3.1.1) f(x) l < ε. In lim x c f(x) = l for some l we say that f has a( finite) limit at c. 34

36 (2) Suppose that E (a, c). We say that f has left limit at c, if for any ε > 0 there exists δ > 0, such that for all x E satisfying c δ < x < c, we have f(x) l < ε. This is denoted by lim f(x) = l. x c (3) Suppose that E (c, b). We say that f has right limit at c, if for any ε > 0 there exists δ > 0, such that for all x E satisfying c < x < c+δ, we have f(x) l < ε. This is denoted by We observe that lim f(x) = l. x c+ Proposition Let f : (a, b) R and c (a, b). It is clear that lim x c f(x) = l is equivalent to both lim x c+ f(x) = l and lim x c f(x) = l. The condition that E contains an interval around c can be replaced by c is an accumulation point of E, see the Appendix, Chapter 3.2. This assumption will ensure that if there is a limit, the limit is unique. Proposition If f : R has a left limit (or has a right limit) at c, this limit is unique. Proof We prove only the case when lim x c f(x) exists. Suppose f has two limits l 1 and l 2 with l 1 l 2. Take ε = 1 4 l 1 l 2 > 0. By definition, there exists δ > 0 such that if 0 < x c < δ, f(x) l 1 < ε, f(x) l 2 < ε. (such x does exists from the assumption that either (a, c) (c, b) E!!) By the triangle inequality, l 1 l 2 f(x) l 1 + f(x) l 2 < 2ε = 1 2 l 1 l 2. This is a contradiction. Example The statement that lim x 0 f(x) = 0 is equivalent to the statement that lim x 0 f(x) = 0. Define g(x) = f(x), then g(x) 0 = f(x). 35

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